Linux 如何在C中初始化二进制信号量
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How to initialise a binary semaphore in C
提问by austinmarton
In the man pageit appears that even if you initialise a semaphore to a value of one:
在手册页中，即使您将信号量初始化为值 1：
sem_init(&mySem, 0, 1);
It could still be incremented to a value greater than 1 with multiple calls to
它仍然可以通过多次调用增加到大于 1 的值
sem_post(&mySem);
But in this code examplethe comment seems to think differently:
但在这个代码示例中，注释似乎有不同的想法：
sem_init(&mutex, 0, 1); /* initialize mutex to 1 - binary semaphore */
Is it possible to initialise a strictly binary semaphore in C?
是否可以在 C 中初始化严格的二进制信号量？
Note: The reason for doing this instead of using a mutex in this case is the sem_post and sem_wait may be called by different threads.
注意：在这种情况下这样做而不是使用互斥锁的原因是 sem_post 和 sem_wait 可能被不同的线程调用。
采纳答案by Dietrich Epp
If you want a strictly binary semaphore on Linux, I suggest building one out of mutexes and condition variables.
如果你想在 Linux 上使用严格的二进制信号量，我建议用互斥体和条件变量构建一个。
struct binary_semaphore {
pthread_mutex_t mutex;
pthread_cond_t cvar;
int v;
};
void mysem_post(struct binary_semaphore *p)
{
pthread_mutex_lock(&p->mutex);
if (p->v == 1)
/* error */
p->v += 1;
pthread_cond_signal(&p->cvar);
pthread_mutex_unlock(&p->mutex);
}
void mysem_wait(struct binar_semaphore *p)
{
pthread_mutex_lock(&p->mutex);
while (!p->v)
pthread_cond_wait(&p->cvar, &p->mutex);
p->v -= 1;
pthread_mutex_unlock(&p->mutex);
}
